Question

For the reaction, $2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)$ $N_2O_5$ disappears at a rate of $x$ moldm$^{-3}$ s$^{-1}$ Find the rate of formation of $O_2$ ?

• A. $x$ moldm$^{-3}$ s$^{-1}$
• B. $2x$ moldm$^{-3}$ s$^{-1}$
• C. $\frac{x}{2}$ moldm$^{-3}$ s$^{-1}$
• D. $\frac{3x}{2}$ moldm$^{-3}$ s$^{-1}$

Hint:

Rate of reaction is the change in concentration divided by its stoichiometric coefficient.

Answer 1

The Correct Option is C

Step 1: Rate Equation
Rate = $-\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{d[O_2]}{dt}$.

Step 2: Analysis
Given disappearance rate of $N_2O_5 = -\frac{d[N_2O_5]}{dt} = x$.

Step 3: Calculation
From step 1: $\frac{d[O_2]}{dt} = \frac{1}{2} \times \left(-\frac{d[N_2O_5]}{dt}\right) = \frac{1}{2}x$.

Step 4: Conclusion
The rate of formation of $O_2$ is $\frac{x}{2}$.
Final Answer: (C)